Download Homework I. Mendelian genetics KEY (Sec 37)

Homework I. Mendelian genetics
KEY (Sec 37)
1- An organism with genotype AaEEDd can make how many different gametes (list
each one)? (1 pt)
It could make 4 different types of gametes: AED, AEd, aED, aED.
2- You are studying tail length in squirrels and perform the standard genetic
crosses. From the following observations, determine how this trait is inherited
(which allele is dominant, and is the trait sex-linked?). Assume that parents are
true breeding. Test your hypothesis with Chi-square. Show all your calculations
for the Chi-square test. (2 pts)
P: short tailed males x long tailed females
F1: 145 long tailed males and 155 short tailed females
(the F2 is the result of crossing a F1 male with a F1 females)
F2: 73 long tailed females, 77 short tailed females, 72 long tailed males and 78
short tailed males.
First, looking at the F1 you see that males and females are different; this observation
ruled out the possibility of this is autosomal, then it is SEX-LINKED. Given the
results in each cross and based on the predictions of Mendelian hypotheses, it can be
hypothesized that Short tail is the dominant trait and it is sex-linked. For this
particular parental cross and given the hypothesis, the expected phenotypic ratios in
the F2 are 1 dominant female : 1 recessive female : 1 dominant male : 1 recessive
male.
Long tailed females
Observed
(O)
73
Expected
(E)
75
Short tailed females
Long tailed males
77
72
Short tailed males
Total
78
300
(O-E)
(O-E)2
X2
-2
4
4 / 75 = 0.053
75
75
2
-3
4
9
4 / 75 = 0.053
9 / 75 = 0.12
75
300
3
9
9 / 75 = 0.12
0.346
You have four phenotypes (long tailed females, long tailed males, short tailed
females, and short tailed males), so your degrees of freedom are 4-1 = 3. The
tabulated Chi-square for 3 degrees of freedom is equal to 7.815. Your calculated Chisquare is less than the theoretical value, which means that you hold your hypothesis
(Short tail is a dominant and sex-linked trait)
3- You are analyzing a dihybrid cross where both mutant traits are autosomal
dominant. You set up the parental cross with each true breeding parent
possessing one wild type trait. You observed 96 mice in the F2 generation. What
is the expected number of each F2 phenotypic group. (1 pts)
Let’s define A = mutant trait 1, a = wild trait 1, B = mutant trait 2, and b = wild trait
2.
For a dihybrid cross, the expected phenotypic ratio is 9 dominant-dominant : 3
dominant-recessive : 3 recessive-dominant : 1 recessive-recessive. Remember that to
get the expected numbers of each phenotype, you have to divide your total
observation (96 mice) by the total number of genotypes (16). The result of this
division is multiplied by each of the expected proportions:
96 / 16 = 6, then:
9 x 6 = 54 mice A_B_
3 x 6 = 18 mice A_bb
3 x 6 = 18 mice aaB_
1 x 6 = 6 mice aabb
4- Both Mrs. Smith and Mrs. Jones had babies the same day in the same hospital.
Mrs. Smith took home a bay girl, whom she named Sharon. Mrs. Jones took
home a girl, whom she named Jane . Mrs. Jones began to suspect, however, that
the child had been accidentally switched with Mrs. Smith baby in the nursery.
Blood test were made; Mr. Smith was type A, Mrs. Smith was type B, Mr. Jones
was type A, Mrs. Jones was type A. Sharon was type O, and Jane was type B.
Had a mix-up occurred? Show your work! (1 pts)
You know the phenotype of each parent but not their genotype. Then, Mr. Smith could
be either IA IA or IA I, and the same thing for Mrs. Smith, Mr. Jones and Mrs. Jones.
We are going to use the following notation:
Mr. Smith
IA __
Mrs. Smith
IB __
Mr. Jones
IA __
Mrs. Jones
IA __
If you do the Punnett square, you will see that:
The Smith baby could be: IA IB, IA i, IB i,or i i
The Jones baby could be: IA IA, IA i, or i i
Therefore, the parents of Jane can only be the Smith; there was a mix -up!
5- You are studying ONE trait that is autosomal recessive. You crossed the
following true breeding parents: autosomal dominant female with an autosomal
recessive male. What are the expected genotypic and phenotypic ratios in the F2
generation? (1 pt)
Let’s define E = dominant and e = recessive. Parents are true breeding that means
they are homozygous:
P: EE x ee (as it is an autosomal trait, it does not matter which is female or male)
e
e
E
Ee
Ee
E
Ee
Ee
Your F1 is 100% Ee that means all the progeny has the dominant phenotype. Now, to
get the expected genotypic and phenotypic ratios in the F2, you breed the F1:
E
e
E
EE
Ee
e
Ee
ee
Genotypic ratio = 1 EE : 2 Ee : 1 ee
Phenotypic ratio = 3 dominant : 1 recessive
6- You have found a blue tomato growing in your garden and wa nt to find its
genetic basis. You make a cross of a true breeding blue tomato plant with true
breeding red tomato plant. Out of 1000 F2 plants you find that 712 are red, and
288 are blue. What is the likely genetic basis for the blue tomato (which trait is
dominant?). Test your hypothesis with a Chi-square test. (2 pts)
The red trait is most likely dominant, and this is probably a monohybrid cross. So,
you have two phenotypic classes: red tomatoes and blue tomatoes. If this trait is
inherited as expected by Mendel’s First Law, you expect to observe 750 red tomatoes
and 250 blue ones in your F2.
Red
Blue
Observed
(O)
712
288
Expected
(E)
750
250
Total
1000
1000
(O-E)
-38
38
(O-E)2
1444
1444
X2
1.92
5.77
7.69
You have only 2 phenotypes, red (dominant) and blue (recessive), so the degrees of
freedom are 2 -1 = 1. The tabulated Chi-square for 1 degree of freedom is equal to
3.841. Your calculated Chi-square is bigger than the theoretical value, which means
that you reject your hypothesis (tomato color is not inherited as expected by first
Mendel’s Laws).